the velocity of bullet becomes halved while travelling through a distance of 6cm. Further distance it has to travel before coming to restalong with procedure plz

Answers (1)
S safeer

Let the incoming velocity of bullet be v, when it travels 6cm the velocity reduces to half. Therefore the acceleration can be obtained as

v^2=u^2-2as \\\text{here u=v and v=v/2 therefore}\\(\frac{v}{2})^2=v^2-2a\times6\\\\\frac{3v^2}{4}=12a\\\\a=\frac{v^2}{16}

When the bullet stops the velocity is zero.

\\v^2=u^2-2as\\\\ 0=\frac{v^2}{4}-2(\frac{v^2}{16})s\\\\s=2cm

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